Here is the graph, you can match it yourself:
Answer:
a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)
Step-by-step explanation:
Let's solve by separating variables:

a) x’=t–sin(t), x(0)=1

Apply integral both sides:

where k is a constant due to integration. With x(0)=1, substitute:

Finally:

b) x’+2x=4; x(0)=5

Completing the integral:

Solving the operator:

Using algebra, it becomes explicit:

With x(0)=5, substitute:

Finally:

c) x’’+4x=0; x(0)=0; x’(0)=1
Let
be the solution for the equation, then:

Substituting these equations in <em>c)</em>

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>
![x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)](https://tex.z-dn.net/?f=x%3De%5E%7B%5Calpha%20t%7D%5BAsin%5Cbeta%20t%2BBcos%5Cbeta%20t%5D%5C%5C%5C%5Cx%3De%5E%7B0%7D%5BAsin%28%282%29t%29%2BBcos%28%282%29t%29%5D%5C%5C%5C%5Cx%3DAsin%28%282%29t%29%2BBcos%28%282%29t%29)
Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

Finally:

Answer:
80/81
Step-by-step explanation:
If a head is twice as likely to occur as a tail, then the probability of getting heads is 2/3 and the probability of getting tails is 1/3.
The probability of getting at least 1 head involves 4 scenarios:
1) 1 Head and 3 Tails
2) 2 Heads and 2 Tails
3) 3 Heads and 1 Tail
4) 4 Heads
Instead of calculate all these scenarios, you could calculate the opposite scenario: 4 Tails. The sum of all possible scenarios is 1, so:
P(at least one head) + P(no heads) = 1
Then, P(at least one head) = 1 - P(no heads)
The probability of 4 tails is:
P(no heads) = P(TTTT) = (1/3)(1/3)(1/3)(1/3)=1/81
Then, P(at least one head) = 1 - 1/81=80/81